# Evaluation of Integral (nln(n))^-1

1. Aug 4, 2012

### jimbobian

1. The problem statement, all variables and given/known data

Evaluate $$\int{\frac{1}{nln(n)} dn }$$

2. Relevant equations

3. The attempt at a solution

I know the answer thanks to WolframAlpha, I just want to understand why my method didn't work.

I took a stab at parts using:
$$u=\frac{1}{ln(n)}$$
$$\frac{dv}{dn}=\frac{1}{n}$$
So this gives:
$$v=ln(n)$$
Using quotient rule:
$$\frac{du}{dn}=\frac{0-(1)(1/n)}{(ln(n))^2}=\frac{-1}{n(ln(n))^2}$$
And therefore as:
$$\int{u\frac{dv}{dn} dn } = uv - \int{v\frac{du}{dn} dn}$$
$$\int{\frac{1}{nln(n)} dn } = ln(n)\frac{1}{ln(n)} - \int{ln(n)\frac{-1}{n(ln(n))^2} dn}$$
$$= 1 + \int{\frac{1}{n ln(n)} dn}$$

Which surely isn't right? Can anyone spot the mistake/reason why this method doesn't work?

Cheers

2. Aug 4, 2012

### uart

Hi Jimbobian. Write the integral as,

$$\int n^{-1} \, \left[ ln(n)\right]^{-1} dn$$

and then note that it is in the form of $\int g' \, f \circ g(n) \, dn$

This is a special form (the product of a function of a function times the derivative of the inner function) that we should always be on the look out for when trying to find an anti-derivative. This form always has a simple solution so long as we know the anti derivative of just the outer function "f()".

3. Aug 4, 2012

### jimbobian

Damn, one month after school finishing and I'd already forgotten to be on a look out for that, despite the number of times my teacher brought it up!

That form would certainly make evaluating it easier (and thank you for spotting it), but I would also like to know why my method didn't work?

4. Aug 4, 2012

### LCKurtz

We pretty much always ignore the constants of integration when we do integration by parts. When you integrate dv we take the constant zero. Your result just says the two antiderivatives differ by a constant, but that doesn't get you anywhere.

I assume you know that integral would be done with a u substitution, letting u = ln(x).

5. Aug 4, 2012

### jimbobian

Thank you! Yes, the actual integration, as uart pointed out, isn't challenging I just got hung up on why this didn't work.

Thank you both, my issue is solved!

6. Aug 4, 2012

### Ray Vickson

If you do a definite integration instead, say from 2 to x, integration by parts just gives you
$$\int_2^x \frac{dn}{n \ln(n)} = \int_2^x \frac{dn}{n \ln(n)},$$
which is no longer contradictory, but is also not useful.

RGV